How do you prove cos (90 deg + A) = -sin A?

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How do you prove cos (90 deg + A) = -sin A?

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Answer 1 · 2016-02-25T19:09:58

see explanation

Explanation:

using $Addition formula$ $color(blue)" Addition formula "$

• cos(A ± B ) = cosAcosB ∓ sinAsinB (Note sign change )

applying this to the question to obtain:

$cos (90^{\circ} + A) = {cos 90}^{\circ} cos A - {sin 90}^{\circ} sin A$ $cos(90^@ + A ) = cos90^@cosA - sin90^@sinA$

you should know the following facts:

${cos 90}^{\circ} = 0 and {sin 90}^{\circ} = 1$ $cos90^@ = 0 " and " sin90^@ = 1$

$\Rightarrow cos (90^{\circ} + A) = 0 \times cos A - 1 \times sin A = 0 - sin A = - sin A$ $rArrcos(90^@ + A ) = 0xxcosA - 1xxsinA = 0-sinA = -sinA$

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How do you prove cos (90 deg + A) = -sin A?

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